Question

A straight copper wire of length $1000$m and cross-sectional area $1.0$ $m{m}^2$ carries a current $4.5$A. Assuming that one free electron corresponds to each copper atom.(Density of copper $=8.96\times {10}^3$ kg $m^{-3}$, Atomic mass of copper $=63.5$g, Resistivity of copper wire$=1.69\times {10}^{-8}\Omega$m).

• A. The time taken by an electron to displace from one end of the wire to the other is $4\times {10}^6$s
• B. The sum of electric force acting on all free electrons in the given wire is $1\times {10}^6$N
• C. The time taken by an electron to displace from one end of the wire to the other is $3\times {10}^6$S
• D. Both (b) and (c)

Answer 1

The correct option is D Both (b) and (c)

Let, the volume density of electrons be $n$

The cuurent flowing through the conductor can be given as:

$I=\rho S_x{V}_d$

$I=ne{S}_x|\overrightarrow{v}|$

$=ne{S}_x\frac{l}{t}$

So, the time taken by the eectron to displace from one end to the other is:

$t=\frac{ne{S}_{x}l}{I}=3\mu s$

(b) The sum of net forces acting on the electrons is:

$=|\left(nv\right)e\overrightarrow{E}|=|nSle\rho \overrightarrow{j}|$. where $\rho$ is resistivity of the material.

$=nS{l}_{\epsilon \rho }\frac{I}{S}=nel\rho I=1.0\mu N$